m = 20*10^6 g = 20*10^3 kg
v = 8000m/3600 s = 2.22 m/s
Ke = (1/2) m v^2 = 10*10^3 * 4.94
= 4.94 * 10^4 Joules
= (1/2) k x^2
x = .4 meter (wow!)
x^2 = .16
(1/2) x^2 = .08
so
k = 4.94 * 10^4/.08 =61.7 *10^4 N/m
= 617,000 N/m
A linear spring of stiffness k is designed to stop the 20-Mg railroad car traveling at 8 km/h within 400mm after impact. Find the smallest value of k that
will produce the desired result.
1 answer